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$MX_2+2AgNO_3\rightarrow M(NO_3)_2+2AgX\downarrow (1)$
$\frac{41,6}{2} 28,7$
$MX_2+Na_2CO_3\rightarrow 2NaX+MCO_3\downarrow (2)$
$\frac{41,6}{2} 19,7$
$\Rightarrow \begin{cases}n_{MX_2 (1)}=\frac{28,7-20,8}{2.108-M} \\n_{MX_2 (2)}=\frac{20,8-19,7}{2.X-60} \end{cases}\Rightarrow \frac{7,9}{216-M}=\frac{1,1}{2X-60}\Rightarrow \begin{cases}M<216 \\X>30\\45,03-\frac{1,1}{15,8}.M=X \end{cases}$
$\Rightarrow 30<X<45,03\Rightarrow X=35,5\Rightarrow X:Cl\Rightarrow M=137\Rightarrow M:Ba$
$\Rightarrow BaCl_2$
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Trả lời 03-02-13 12:09 AM
viet130480
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